Problem: $h(x)=\begin{cases} 2x+1&\text{for }x<-2 \\\\ \dfrac{1}{x-1}&\text{for }-2\leq x<1 \end{cases}$ Find $\lim_{x\to -3}h(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-5$ (Choice B) B $-3$ (Choice C) C $-\dfrac14$ (Choice D) D The limit doesn't exist.
Solution: Let's find the limit as $x$ approaches $-3$. We will use the fact that $h(x)=2x+1$ for $x$ -values smaller than $-2$. $\begin{aligned} &\phantom{=}\lim_{x\to -3}h(x) \\\\ &=\lim_{x\to -3}2x+1 \\\\ &=2(-3)+1&\gray{\text{Direct substitution}} \\\\ &=-5 \end{aligned}$ In conclusion, we found that $\lim_{x\to -3}h(x)=-5$.